Closing "Fibonacci's Hole"
Oh gosh, I spent ages with it. On the pictures you posted it looks particularly nice because of lines are actually spirals. If they were straight it won't look nice. On your picture pole doesn't have quads and also has a lot of adjacent edges. So if you don't want this, maybe you can just stop (I would stop earlier) and connect the rest of the points using a different triangulation method?

In your implementation there is same distance of fib between each pair of vertices you connect. So the question comes when index + fib is larger than number of vertices. Where do you want to connect the rest of them?

This is what I get for k = 7. 


And this for k = 9.



This is the bit I changed. But I am no expert in either bmesh nor bpython, so maybe you can polish it.

# make mesh using 9th and 10th fibonacci numbers       
fibk = 7
bm = sfmesh.bm
bm.verts.ensure_lookup_table()
for i in [fibk, fibk + 1]:
    fib = F(i)
    print("Fib number ", i, fib)
    for k in range(-fib+1, n-1 ):
        for i in [k, min(n-1,k + fib)]:           
            try:
                bm.edges.new([bm.verts[max(0,k)],bm.verts[min(n-1,k + fib)]] )      

            except(ValueError):
                print('duplicate')


1) Also you can find code  here which produces this, maybe it will be useful. 


2) Also look at this post which leads to this solution:


It seems there are in general not many ways to make a nice quad mesh of a sphere.No matter how you do it, there always is going to be a seem. I think 1) and 2) two are basically the main solutions.

Bo, if you want Fibonacci I would go with 2),so cut it off and close it as they did, if you want regular - with 1).